## Wednesday, September 29, 2010

### Exam I Feedback for Students

I just completed the grading for exam I and I wanted to summarize my feedback here on the blog. Hopefully this will allow you to look at these issues with your textbook in hand. I will also mention them during class- we can do some examples as a class.

Overall the average was in the mid 60s but this is because there were a few really low scores that really pulled it down. There were 6 people who scored above 90%. One person got a 98%. So- it was a manageable test. I had a good number of people in the 80 range and a few in the 70s. There were also a good chunk below 70 and this concerns me. My goal is to work hard with each of you that scored in this range so you can bring up your scores. Your homework/quiz/activity grade will help you bring your score up somewhat but you need to concentrate on performance on the tests to really do well in the class (A or B grade).

Several of you still struggle with the concepts of conversion, units and volume/density. There are a few of you that only struggled with this concept- if you had answered these questions correctly you would have been above 90% on the test. So..... keep these things in mind: Zeroes that occur before the decimal place or after the decimal place but do not follow a digit that is 1-9 are not considered significant. Therefore the number 0.00347 has only 3 significant digits. The three digits preceding the "347" are not considered significant. The number 0.03470 has 4 significant digits. The last zero is considered significant because although it falls on the right of the decimal, it follows a digit that is 1-9 and is therefore considered significant. Everybody needs to keep in mind that I consider it crucial for you to show your calculations and logic in the short answer section. I CANNOT award partial credit for an incorrect answer if you do not carefully show me your mathematics progression. I wanted problem 44 and 45 to be an easy 20 points for everyone. These were the density/mass/volume conversions and the volume conversion problems. Several of you got an answer that was on the right track but you did not receive partial credit because you did not show your calculations.
Keep in mind that while a cm3 (superscript 3) is equal to a mL, it is a dm3 (superscript 3) that is equal to a Liter. There are two ways to solve this problem (convert to dm3 or just convert from mL to Liter with the conversion 1000 mL/L) Several of you need to review your basic conversion factors- I gave you a table of all of the factors but people did not use this information correctly. I had people converting cm into meters with the conversion factor 0.01cm/m. It is opposite: 100 cm/meter. Review how these are used in word problems.
I wanted people to notice how I put a problem on the test that was an example we worked in class. Actually- this problem was in both the multiple choice section and the calculation section: You were supposed to figure out the relative atomic mass of an element based on two isotopes found in nature. There were too many people who missed this problem. I was hoping you review lecture notes and examples from class so you can get these types of problems correct.

I was very pleased with the knowledge of the class about atomic theory. The last problem was supposed to be an easy 5 points for everyone. Most people got 4/5. The two modifications to Dalton in Modern Atomic Theory are as follows:  1. Discovery of neutrons allows atoms of the same element to have different masses 2. Atoms are not the smallest building blocks of nature due to the discovery of subatomic particles. (Some of you stated that H2O and H2O2 prove that the law of definite proportions is not true and this is false. The law of definite proportions just states that compounds always contain the same ratio of elements. It doesn't mean that the ratios can't vary between compounds  (which they do only in whole number values))

Problem number 10 was a commonly missed question dealing with the law of definite proportions. The ratio of sulfur to oxygen in SO3 is 32:48.  This is obtained by multiplying the atomic mass of sulfur ( 32) by 1 (for one atom of sulfur)  (from the PT) and the atomic mass of oxygen (16) by 3 (from the PT). The mass ratio of sulfur to oxygen in any sample of SO3 will always be 32:48 by the law of definite proportions.

I'll post the key and talk more about it in class.